Transcript Physics 207: Lecture 2 Notes

Lecture 10

Goals:
 Exploit Newton’s 3rd Law in problems with friction
 Employ Newton’s Laws in 2D problems with circular motion
Assignment: HW5, (Chapter 7, due 2/24, Wednesday)
For Tuesday: Finish reading Chapter 8, First four sections in
Chapter 9
Physics 207: Lecture 10, Pg 1
Example: Friction and Motion

A box of mass m1 = 1 kg is being pulled by a horizontal
string having tension T = 40 N. It slides with friction
(mk= 0.5) on top of a second box having mass m2 = 2 kg,
which in turn slides on a smooth (frictionless) surface.
 What is the acceleration of the bottom box ?
Key Question: What is the force on mass 2 from mass 1?
v
T
a=?
m1
m2
slides with friction (mk=0.5 )
slides without friction
Physics 207: Lecture 10, Pg 2
Example
Solution

First draw FBD of the top box:
v
N1
T
fk = mKN1 = mKm1g
m1
m1g
Physics 207: Lecture 10, Pg 3
Example
Solution

Newtons 3rd law says the force box 2 exerts on box 1 is
equal and opposite to the force box 1 exerts on box 2.

As we just saw, this force is due to friction:
Reaction
f2,1 = -f1,2
Action
m1
f1,2 = mKm1g = 5 N
m2
Physics 207: Lecture 10, Pg 4
Example
Solution

Now consider the FBD of box 2:
N2
N1 = m1g
f2,1 = mkm1g
m2
m2g
Physics 207: Lecture 10, Pg 5
Example
Solution

Finally, solve Fx = max in the horizontal direction:
mK m1g = m2ax
m1m k g 5 N
ax 


m2
2 kg
= 2.5 m/s2
f2,1 = mKm1g
m2
Physics 207: Lecture 10, Pg 6
Home Exercise
Friction and Motion, Replay
A box of mass m1 = 1 kg, initially at rest, is now pulled by a
horizontal string having tension T = 10 N. This box (1) is on top
of a second box of mass m2 = 2 kg. The static and kinetic
coefficients of friction between the 2 boxes are ms=1.5 and mk=
0.5. The second box can slide freely (frictionless) on an smooth
surface.
Question:
Compare the acceleration of box 1 to the acceleration of box 2?

a1
T
m1
a2
m2
friction coefficients
ms=1.5 and mk=0.5
slides without friction
Physics 207: Lecture 10, Pg 7
Exercise Tension example
Compare the strings below in settings (a) and (b) and their
tensions.
A.
B.
C.
D.
Ta = ½ Tb
Ta = 2 Tb
Ta = Tb
Correct answer is not given
Physics 207: Lecture 10, Pg 10
Problem 7.34 Hint (HW 6)
Suggested Steps
 Two independent free body diagrams are necessary
 Draw in the forces on the top and bottom blocks
 Top Block
 Forces: 1. normal to bottom block 2. weight 3. rope tension
and 4. friction with bottom block (model with sliding)
 Bottom Block
 Forces:
1. normal to bottom surface
2. normal to top block interface
3. rope tension (to the left)
4. weight (2 kg)
5. friction with top block
6. friction with surface
7. 20 N
Use Newton's 3rd Law to deal with the force pairs
(horizontal & vertical) between the top and bottom block.
Physics 207: Lecture 10, Pg 11
On to Chapter8
Reprisal of : Uniform Circular Motion
For an object moving along a curved trajectory with constant
speed
a = ar (radial only)
v
vt2
|ar |=
r
ar
Physics 207: Lecture 10, Pg 12
Non-uniform Circular Motion
For an object moving along a curved trajectory,
with non-uniform speed
a = ar + at (radial and tangential)
at
vT2
|ar |=
r
ar
|at |=
d| v |
dt
Physics 207: Lecture 10, Pg 13
Key steps
 Identify
forces (i.e., a FBD)
 Identify
axis of rotation
 Apply
conditions (position, velocity &
acceleration)
Physics 207: Lecture 10, Pg 14
Example
The pendulum
axis of rotation
Consider a person on a swing:
When is the tension on the rope largest?
And at that point is it :
(A) greater than
(B) the same as
(C) less than
the force due to gravity acting on the person?
Physics 207: Lecture 10, Pg 15
Example
Gravity, Normal Forces etc.
axis of rotation
T
T
y
vT
q
x
mg
at top of swing vT = 0
mg
at bottom of swing vT is max
Fr = m 02 / r = 0 = T – mg cos q
T = mg cos q
Fr = m ac = m vT2 / r = T - mg
T = mg + m vT2 / r
T < mg
T > mg
Physics 207: Lecture 10, Pg 16
Conical Pendulum (very different)

Swinging a ball on a string of length L around your head
(r = L sin q)
axis of rotation
S Fr = mar = T sin q
L
S Fz = 0 = T cos q– mg
so
T = mg / cos q (> mg)
r
mar = (mg / cos q) (sin q )
ar = g tan qvT2/r
 vT = (gr tan q)½
Period:
t = 2p r / vT =2p (r cot q/g)½
= 2p (L cos q/g)½
Physics 207: Lecture 10, Pg 17
Conical Pendulum (very different)

Swinging a ball on a string of length L around your head
axis of rotation
Period:
t = 2p r / vT =2p (r cot q/g)½
= 2p (L cos q/ g )½
= 2p (5 cos 5/ 9.8 )½ = 4.38 s
= 2p (5 cos 10/ 9.8 )½ = 4.36 s
= 2p (5 cos 15/ 9.8 )½ = 4.32 s
L
r
Physics 207: Lecture 10, Pg 18
Another example of circular motion
Loop-the-loop 1
A match box car is going to do a loop-the-loop
of radius r.
What must be its minimum speed vt at the top
so that it can manage the loop successfully ?
Physics 207: Lecture 10, Pg 19
Loop-the-loop 1
To navigate the top of the circle its tangential
velocity vT must be such that its centripetal
acceleration at least equals the force due to
gravity. At this point N, the normal force,
goes to zero (just touching).
Fr = mar = mg = mvT2/r
vT
vT = (gr)1/2
mg
Physics 207: Lecture 10, Pg 20
Loop-the-loop 2
The match box car is going to do a loop-the-loop.
If the speed at the bottom is vB, what is the
normal force, N, at that point?
Hint: The car is constrained to the track.
Fr = mar = mvB2/r = N - mg
N = mvB2/r + mg
N
v
mg
Physics 207: Lecture 10, Pg 21
Loop-the-loop 3
Once again the car is going to execute a loop-the-loop.
What must be its minimum speed at the bottom so
that it can make the loop successfully?
This is a difficult problem to solve using just forces. We
will skip it now and revisit it using energy
considerations later on…
Physics 207: Lecture 10, Pg 22
Example, Circular Motion Forces with Friction
(recall mar = m |vT | 2 / r Ff ≤ ms N )

How fast can the race car go?
(How fast can it round a corner with this radius of curvature?)
mcar= 1600 kg
mS = 0.5 for tire/road
r = 80 m
g = 10 m/s2
r
Physics 207: Lecture 10, Pg 24
Example

Only one force is in the horizontal direction: static friction
x-dir: Fr = mar = -m |vT
y-dir: ma = 0 = N – mg
|2/
y
r = Fs = -ms N (at maximum)
N = mg
vT = (ms m g r / m )1/2
vT = (ms g r )1/2 = (0.5x 10 x 80)1/2
vT = 20 m/s
N
Fs
mg
mcar= 1600 kg
mS = 0.5 for tire/road
r = 80 m
g = 10 m/s2
Physics 207: Lecture 10, Pg 25
x
Another Example
A horizontal disk is initially at rest and very slowly undergoes
constant angular acceleration. A 2 kg puck is located a point
0.5 m away from the axis. At what angular velocity does it slip
(assuming aT << ar at that time) if ms=0.8 ?
 Only one force is in the horizontal direction: static friction
y
x-dir: Fr = mar = -m |vT | 2 / r = Fs = -ms N (at w)

y-dir: ma = 0 = N – mg
N = mg
vT = (ms m g r / m )1/2
vT = (ms g r )1/2 = (0.8x 10 x 0.5)1/2
vT = 2 m/s  w = vT / r = 4 rad/s
N
Fs
mg
mpuck= 2 kg
mS = 0.8
r = 0.5 m
g = 10 m/s2
Physics 207: Lecture 10, Pg 26
x
Banked Curves
In the previous car scenario, we drew the following free
body diagram for a race car going around a curve on a
flat track.
n
Ff
mg
What differs on a banked curve?
Physics 207: Lecture 10, Pg 27
Lecture 10
Assignment: HW5, (Chapters 7, due 2/24, Wednesday)
For Tuesday: Finish reading Chapter 8 and start Chapter 9
Physics 207: Lecture 10, Pg 28